How much of each substance do I need if I have #"100.0 g"# of iron?

My chemistry book says:

"Aluminium powder and iron (III) oxide react with each other in a "thermite" reaction, which releases a lot of heat. This reaction has been used (for example) in building railways, which requires molten iron. The chemical equation regarding this reaction is 2 Al(s) + Fe2O3(s) --> 2 Fe(l) + Al2O3(s).

How much of each substance is needed, if you have 100,0 grams of iron?"

I'm sorry for possible mistakes in my grammar, english is not my first language.

1 Answer
Aug 22, 2017


Here's what I got.


You know that the balanced chemical equation that describes this reaction looks like this

#2"Al"_ ((s)) + "Fe"_ 2"O"_ (3(s)) -> 2"Fe"_ ((l)) + "Al"_ 2"O"_ (3(s))#

Take a look at the coefficients added in front of the four chemical species that are involved in the reaction--keep in mind that if no coefficient is visible, that means that the coefficient is actually equal to #1#.

In this case, you have a #2# in front of aluminium and a #1# in front of iron(III) oxide on the reactants' side and a #2# in front of iron and a #1# in front of aluminium oxide on the products' side.

This tells you that the reaction consumes #2# moles of aluminium and #1# mole of iron(III) oxide and produces #2# moles of iron and #1# mole of aluminium oxide.

The problem tells you that the reaction produced #"100.0 g"# of iron, so start by converting the mass of iron to moles. To do that, use the molar mass of iron

#100.0 color(red)(cancel(color(black)("g"))) * "1 mole Fe"/(55.845color(red)(cancel(color(black)("g")))) = "1.791 moles Fe"#

Now, use the aforementioned mole ratios to find the number of moles of aluminium and of iron(III) oxide needed to produce that many moles of iron

#1.791 color(red)(cancel(color(black)("moles Fe"))) * ("1 mole Fe"_2"O"_3)/(2color(red)(cancel(color(black)("moles Fe")))) = "0.8955 moles Fe"_2"O"_3#

#1.791 color(red)(cancel(color(black)("moles Fe"))) * "2 moles Al"/(2color(red)(cancel(color(black)("moles Fe")))) = "1.791 moles Al"#

To convert the number of moles to grams, use the molar masses of the two reactants

#0.8955 color(red)(cancel(color(black)("moles Fe"_2"O"_3))) * "159.69 g"/(1color(red)(cancel(color(black)("mole Fe"_2"O"_3)))) = color(darkgreen)(ul(color(black)("143.0 g Fe"_2"O"_3)))#

#1.791 color(red)(cancel(color(black)("moles Al"))) * "26.98 g"/(1color(red)(cancel(color(black)("mole Al")))) = color(darkgreen)(ul(color(black)("48.32 g")))#

The answers are rounded to four sig figs, the number of sig figs you have for the mass of iron.